Question

In a NEUTRAL solution, how does `[HO^-]` compare to `[H_3O^+]`?

Answer 1

How do you think it compares...........?

Explanation:

In aqueous solution under standard conditions, the following equilibrium operates:

`2H_2OrightleftharpoonsH_3O^+ + HO^-`

At `298*K`, the ion-product, `[H_3O^+][HO^-]=10^-14`,

we know that `[H_3O^+]=[HO^-]` by specification (we have a neutral solution), and thus,

`[H_3O^+]=[HO^-]=sqrt(10^-14)*mol*L^-1=10^-7*mol*L^-1`.